Proving that $2BbbN$ is countable Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is a surjection from the natural numbers enough to show that a set is countable?Proving : Every infinite subset of countable set is countableProblems regarding countable sets.Proving Every open set in $Bbb R$ is a countable union of open intervals.All distances are rational prove the set is countable$Bbb N times Bbb N$ is countable inductionProving that $Bbb Q$ is dense in $Bbb R$.Proof that finite union of countable sets is countableHelp proving there is a sequence of rational numbersProving a set to be countableCountability as bijective correspondence
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Proving that $2BbbN$ is countable
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is a surjection from the natural numbers enough to show that a set is countable?Proving : Every infinite subset of countable set is countableProblems regarding countable sets.Proving Every open set in $Bbb R$ is a countable union of open intervals.All distances are rational prove the set is countable$Bbb N times Bbb N$ is countable inductionProving that $Bbb Q$ is dense in $Bbb R$.Proof that finite union of countable sets is countableHelp proving there is a sequence of rational numbersProving a set to be countableCountability as bijective correspondence
$begingroup$
To prove that $2BbbN$ is countable, is it sufficient to define a bijection $f:Bbb2Nrightarrow BbbN$ given by $f(n)=fracn2, forall nin Bbb2N$?
real-analysis analysis
edited Mar 27 at 6:04
Eevee Trainer
10.6k31842
asked Mar 27 at 5:57
Ali LodhiAli Lodhi
413
$endgroup$
add a comment |
$begingroup$
To prove that $2BbbN$ is countable, is it sufficient to define a bijection $f:Bbb2Nrightarrow BbbN$ given by $f(n)=fracn2, forall nin Bbb2N$?
real-analysis analysis
edited Mar 27 at 6:04
Eevee Trainer
10.6k31842
asked Mar 27 at 5:57
Ali LodhiAli Lodhi
413
$endgroup$
5
$begingroup$
Yes, it is sufficient.
$endgroup$
– Kavi Rama Murthy
Mar 27 at 5:58
add a comment |
$begingroup$
To prove that $2BbbN$ is countable, is it sufficient to define a bijection $f:Bbb2Nrightarrow BbbN$ given by $f(n)=fracn2, forall nin Bbb2N$?
real-analysis analysis
edited Mar 27 at 6:04
Eevee Trainer
10.6k31842
asked Mar 27 at 5:57
Ali LodhiAli Lodhi
413
$endgroup$
To prove that $2BbbN$ is countable, is it sufficient to define a bijection $f:Bbb2Nrightarrow BbbN$ given by $f(n)=fracn2, forall nin Bbb2N$?
real-analysis analysis
real-analysis analysis
edited Mar 27 at 6:04
Eevee Trainer
10.6k31842
asked Mar 27 at 5:57
Ali LodhiAli Lodhi
413
edited Mar 27 at 6:04
Eevee Trainer
10.6k31842
asked Mar 27 at 5:57
Ali LodhiAli Lodhi
413
edited Mar 27 at 6:04
Eevee Trainer
10.6k31842
edited Mar 27 at 6:04
Eevee Trainer
10.6k31842
edited Mar 27 at 6:04
Eevee Trainer
10.6k31842
10.6k31842
asked Mar 27 at 5:57
Ali LodhiAli Lodhi
413
asked Mar 27 at 5:57
Ali LodhiAli Lodhi
413
asked Mar 27 at 5:57
Ali LodhiAli Lodhi
413
413
5
$begingroup$
Yes, it is sufficient.
$endgroup$
– Kavi Rama Murthy
Mar 27 at 5:58
add a comment |
5
$begingroup$
Yes, it is sufficient.
$endgroup$
– Kavi Rama Murthy
Mar 27 at 5:58
5
5
$begingroup$
Yes, it is sufficient.
$endgroup$
– Kavi Rama Murthy
Mar 27 at 5:58
$begingroup$
Yes, it is sufficient.
$endgroup$
– Kavi Rama Murthy
Mar 27 at 5:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is sufficient. Of course, you might want a formal proof that it is a bijection, which is quite easy: $n/2=x$ implies $n=2x$ (injectivity) and for any $xinmathbb N$ it suffices to take $n=2xin 2mathbb N$ (surjectivity). In general, the map $mathbb Nto amathbb N$ given by $xmapsto ax$ gives a bijection between these two sets, so $amathbb N$ is always countable (for any $ainmathbb C$).
answered Mar 27 at 6:03
YiFanYiFan
5,4032828
$endgroup$
add a comment |
$begingroup$
Note:
To prove that $S= 2mathbbN $ countable it suffices to show :
There exists a surjection $f: mathbbN rightarrow S$.
Choose $f(n)=2n$.
Is a surjection from the natural numbers enough to show that a set is countable?
answered Mar 27 at 7:51
Peter SzilasPeter Szilas
12k2822
$endgroup$
$begingroup$
The $n/2$ approach is backwards. This answer, the $2n$ approach is right. Think of it this way...you want to COUNT them. This means you answer the questions, which is the first, second, etc., and then you need to show you didn't miss any (surjectivity).
$endgroup$
– Dean C Wills
Mar 27 at 15:53
1
$begingroup$
Dean.Thanks for your comment:)
$endgroup$
– Peter Szilas
Mar 27 at 17:11
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
It is sufficient. Of course, you might want a formal proof that it is a bijection, which is quite easy: $n/2=x$ implies $n=2x$ (injectivity) and for any $xinmathbb N$ it suffices to take $n=2xin 2mathbb N$ (surjectivity). In general, the map $mathbb Nto amathbb N$ given by $xmapsto ax$ gives a bijection between these two sets, so $amathbb N$ is always countable (for any $ainmathbb C$).
answered Mar 27 at 6:03
YiFanYiFan
5,4032828
$endgroup$
add a comment |
$begingroup$
It is sufficient. Of course, you might want a formal proof that it is a bijection, which is quite easy: $n/2=x$ implies $n=2x$ (injectivity) and for any $xinmathbb N$ it suffices to take $n=2xin 2mathbb N$ (surjectivity). In general, the map $mathbb Nto amathbb N$ given by $xmapsto ax$ gives a bijection between these two sets, so $amathbb N$ is always countable (for any $ainmathbb C$).
answered Mar 27 at 6:03
YiFanYiFan
5,4032828
$endgroup$
add a comment |
$begingroup$
It is sufficient. Of course, you might want a formal proof that it is a bijection, which is quite easy: $n/2=x$ implies $n=2x$ (injectivity) and for any $xinmathbb N$ it suffices to take $n=2xin 2mathbb N$ (surjectivity). In general, the map $mathbb Nto amathbb N$ given by $xmapsto ax$ gives a bijection between these two sets, so $amathbb N$ is always countable (for any $ainmathbb C$).
answered Mar 27 at 6:03
YiFanYiFan
5,4032828
$endgroup$
It is sufficient. Of course, you might want a formal proof that it is a bijection, which is quite easy: $n/2=x$ implies $n=2x$ (injectivity) and for any $xinmathbb N$ it suffices to take $n=2xin 2mathbb N$ (surjectivity). In general, the map $mathbb Nto amathbb N$ given by $xmapsto ax$ gives a bijection between these two sets, so $amathbb N$ is always countable (for any $ainmathbb C$).
answered Mar 27 at 6:03
YiFanYiFan
5,4032828
answered Mar 27 at 6:03
YiFanYiFan
5,4032828
answered Mar 27 at 6:03
YiFanYiFan
5,4032828
answered Mar 27 at 6:03
YiFanYiFan
5,4032828
5,4032828
add a comment |
add a comment |
$begingroup$
Note:
To prove that $S= 2mathbbN $ countable it suffices to show :
There exists a surjection $f: mathbbN rightarrow S$.
Choose $f(n)=2n$.
Is a surjection from the natural numbers enough to show that a set is countable?
answered Mar 27 at 7:51
Peter SzilasPeter Szilas
12k2822
$endgroup$
$begingroup$
The $n/2$ approach is backwards. This answer, the $2n$ approach is right. Think of it this way...you want to COUNT them. This means you answer the questions, which is the first, second, etc., and then you need to show you didn't miss any (surjectivity).
$endgroup$
– Dean C Wills
Mar 27 at 15:53
1
$begingroup$
Dean.Thanks for your comment:)
$endgroup$
– Peter Szilas
Mar 27 at 17:11
add a comment |
$begingroup$
Note:
To prove that $S= 2mathbbN $ countable it suffices to show :
There exists a surjection $f: mathbbN rightarrow S$.
Choose $f(n)=2n$.
Is a surjection from the natural numbers enough to show that a set is countable?
answered Mar 27 at 7:51
Peter SzilasPeter Szilas
12k2822
$endgroup$
$begingroup$
The $n/2$ approach is backwards. This answer, the $2n$ approach is right. Think of it this way...you want to COUNT them. This means you answer the questions, which is the first, second, etc., and then you need to show you didn't miss any (surjectivity).
$endgroup$
– Dean C Wills
Mar 27 at 15:53
1
$begingroup$
Dean.Thanks for your comment:)
$endgroup$
– Peter Szilas
Mar 27 at 17:11
add a comment |
$begingroup$
Note:
To prove that $S= 2mathbbN $ countable it suffices to show :
There exists a surjection $f: mathbbN rightarrow S$.
Choose $f(n)=2n$.
Is a surjection from the natural numbers enough to show that a set is countable?
answered Mar 27 at 7:51
Peter SzilasPeter Szilas
12k2822
$endgroup$
Note:
To prove that $S= 2mathbbN $ countable it suffices to show :
There exists a surjection $f: mathbbN rightarrow S$.
Choose $f(n)=2n$.
Is a surjection from the natural numbers enough to show that a set is countable?
answered Mar 27 at 7:51
Peter SzilasPeter Szilas
12k2822
answered Mar 27 at 7:51
Peter SzilasPeter Szilas
12k2822
answered Mar 27 at 7:51
Peter SzilasPeter Szilas
12k2822
answered Mar 27 at 7:51
Peter SzilasPeter Szilas
12k2822
12k2822
$begingroup$
The $n/2$ approach is backwards. This answer, the $2n$ approach is right. Think of it this way...you want to COUNT them. This means you answer the questions, which is the first, second, etc., and then you need to show you didn't miss any (surjectivity).
$endgroup$
– Dean C Wills
Mar 27 at 15:53
1
$begingroup$
Dean.Thanks for your comment:)
$endgroup$
– Peter Szilas
Mar 27 at 17:11
add a comment |
$begingroup$
The $n/2$ approach is backwards. This answer, the $2n$ approach is right. Think of it this way...you want to COUNT them. This means you answer the questions, which is the first, second, etc., and then you need to show you didn't miss any (surjectivity).
$endgroup$
– Dean C Wills
Mar 27 at 15:53
1
$begingroup$
Dean.Thanks for your comment:)
$endgroup$
– Peter Szilas
Mar 27 at 17:11
$begingroup$
The $n/2$ approach is backwards. This answer, the $2n$ approach is right. Think of it this way...you want to COUNT them. This means you answer the questions, which is the first, second, etc., and then you need to show you didn't miss any (surjectivity).
$endgroup$
– Dean C Wills
Mar 27 at 15:53
$begingroup$
The $n/2$ approach is backwards. This answer, the $2n$ approach is right. Think of it this way...you want to COUNT them. This means you answer the questions, which is the first, second, etc., and then you need to show you didn't miss any (surjectivity).
$endgroup$
– Dean C Wills
Mar 27 at 15:53
1
1
$begingroup$
Dean.Thanks for your comment:)
$endgroup$
– Peter Szilas
Mar 27 at 17:11
$begingroup$
Dean.Thanks for your comment:)
$endgroup$
– Peter Szilas
Mar 27 at 17:11
add a comment |
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Yes, it is sufficient.
$endgroup$
– Kavi Rama Murthy
Mar 27 at 5:58